anonymous
  • anonymous
multiplying radicals. please help.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[(\sqrt{2}-\sqrt{-2}) (\sqrt{8}+\sqrt{-2})\]
anonymous
  • anonymous
Use the FOIL method just like (x+2)(x+2)
anonymous
  • anonymous
\[(\sqrt{2}-i \sqrt{2}) (2\sqrt{2}+i \sqrt{2})\] . . . \[(2\sqrt{2}+\sqrt{2})+i(2\sqrt{2}-2\sqrt{2})\]

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anonymous
  • anonymous
i am stuck right there.. because in my mind the i part cancels out.. but i know it is wrong because thats not the answer in the book
anonymous
  • anonymous
\( (\sqrt{2}-\sqrt{-2}) (\sqrt{8}+\sqrt{-2}) \) FOIL F = First O = Outer I = Inner L = Last F \( (\sqrt{2})(\sqrt{8}) = \sqrt{16} = 4 \) O \( (\sqrt{2})(\sqrt{-2}) = 2i \) Do you see what I am doing here?
anonymous
  • anonymous
\( (\sqrt{2}-\sqrt{-2}) (\sqrt{8}+\sqrt{-2}) \) FOIL F = First O = Outer I = Inner L = Last F \( (\sqrt{2})(\sqrt{8}) = \sqrt{16} = 4 \) O \( (\sqrt{2})(\sqrt{-2}) = 2i \) I \( (-\sqrt{2})(\sqrt{-8}) = -4i \) L \( (-\sqrt{2})(\sqrt{-2}) = \sqrt{-4} = 2 \) Do you see what I am doing? Can you put the rest together? It is almost done.
anonymous
  • anonymous
And you left.....
anonymous
  • anonymous
sorry for the late response. it ll be 6-2i
anonymous
  • anonymous
thank you that makes more sense.
anonymous
  • anonymous
Good job correct!!!

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